 Solution

A counterexample can even use a nonalgebraic isolated type. Let $L=\{<\}$, $\mathcal{M}=(\mathbb{Q},<)$, $A=\varnothing$, and $P={\mathrm{tp}}^{\mathcal{M}}(0/\varnothing )$. The quantifier elimination for dense linear orders without endpoints shows that all elements have the same complete type over the empty set: a one-variable quantifier-free formula has no order information except $x=x$ and $x<x$.

Thus $x=x$ isolates $P$. Every $\mathcal{N}\equiv \mathcal{M}$ is a dense linear order without endpoints, and every element of $N$ realizes $P$. Such a model cannot omit it. Even a type with infinitely many realizations need not be omittable; isolation is the relevant obstruction.